Ask question

Ask question

All categories

1 year ago

15

If a total force exerted by water in a container with a bottom area of 3 square meters is 900 newtons, what's the water pressure

at the bottom of the container?

1 answer:

anyanavicka [17]1 year ago

6 0

<h3>Answer:</h3>

p=300 Pa

<h3>Explanation:</h3>

_______________

S=3 m²

F=900 N

_______________

p - ?

_______________

p=F/S=900 N / 3 m² = 300 Pa

You might be interested in

How can I shorten this and for it to sill make sense?

Julli [10]

Answer:

Radiation is the emission or transmission of energy in the form of waves.

Explanation:

.

8 0

3 years ago

Read 2 more answers

sukhopar [10]

Answer:

<u>: WHY DIDN'T THE POD DOCK LIKE IT WAS SUPPOSED TO DO?</u><u> </u>

<u>ANSWER</u><u>;</u>

The force exerted by the thrusters caused the pod to change direction.

WHAT NEW THEORIES DO YOU HAVE?

ANSWER;

This pod moved differently because it was more massive.

<em><u>C</u></em><em><u>A</u></em><em><u>R</u></em><em><u>R</u></em><em><u>Y</u></em><em><u>O</u></em><em><u>N</u></em><em><u>L</u></em><em><u>E</u></em><em><u>A</u></em><em><u>R</u></em><em><u>N</u></em><em><u>I</u></em><em><u>N</u></em><em><u>G</u></em><em><u>:</u></em><em><u>)</u></em>

3 0

2 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago

A positive test charge of 8.5 × 10 negative 7 Columbus experiences a force of 4.1 × 10 negative 1 N calculate the electric field

Sophie [7]

Explanation:

direction of electric field is same as that of force experienced by the test charge

8 0

3 years ago

Like his parents, Thomas is short, stocky, and heavyset. Although he exercises regularly and eats a well-balanced meal, his body

Viktor [21]

Answer:

Genetics?

Explanation:

8 0

3 years ago

Read 2 more answers